Question

A sample of gaseous PCl5 was introduced into an evacuated flask so that the pressure of pure PCl5 would be 0.50 atm at 523 K. However, PCl5 decomposes to gaseous PCl3 and Cl2, and the actual pressure in the flask was found to be 0.78 atm. Calculate Kp for the decomposition reaction PCl5(g) PCl3(g) Cl2(g) at 523 K. Also, calculate K at this temperature.

Answer 1

Answer:

Kp = 0.35 & Kc = 0.0075

Explanation:

                                PCl₅ ⇄ PCl₃ + Cl₂

Initial(atm)                 0.5         0       0

At equilibrium         0.5 - p      p       p

• Total pressure at equilibrium = 0.78
• 0.5 - p + p + p = 0.78
• p = 0.28

So partial pressure of PCl₅ = 0.5 - 0.28 = 0.22 atm

Similarly PCl₃ = 0.28 atm & Cl₂ = 0.28 atm

According to law of mass action

     

•                  Kp     =       $\frac{(Ppcl3)(Pcl2)}{(Ppcl5)}$
•                  Kp     =     $\frac{(0.28)(0.28)}{(0.22)}$
•                  Kp     =     0.35

We know Kp = Kc x (RT)Δng

               Δng = change in gaseous moles

                       = product moles - reactant moles

                       = 2 - 1

                        = 1

R = universal gas constant = 0.0821 lit. atm. mol⁻¹. K⁻¹

 So Kc = $\frac{(0.35) }{(0.0821 )(523 K)}$

            = 0.0081